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If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the

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If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink] New post 16 Oct 2011, 03:07
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If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the value of y ?

1. x + y = 3
2. x - y = -3

(C) 2008 GMAT Club - m02#25

Hi All-

I'm struggling to see the logic behind the OA in this question. Taking the information from the question stem, we get y= (x+y)(x-y)/ (x-y), therefore y=x+y. Its easy to see that statement 1 is sufficient, but for statement 2, we will still get y=x+y because y=-3(x+y)/-3. We can pick a number of different values for x-y that equal -3 that fit the question stem (x is not equal to y). Can someone with better math skills than myself explain why statement 2 is sufficient? Cheers!

-Ken
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Re: Tricky Algebra [#permalink] New post 16 Oct 2011, 03:14
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kenman wrote:
If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the value of y ?

1. x + y = 3
2. x - y = -3

(C) 2008 GMAT Club - m02#25

Hi All-

I'm struggling to see the logic behind the OA in this question. Taking the information from the question stem, we get y= (x+y)(x-y)/ (x-y), therefore y=x+y. Its easy to see that statement 1 is sufficient, but for statement 2, we will still get y=x+y because y=-3(x+y)/-3. We can pick a number of different values for x-y that equal -3 that fit the question stem (x is not equal to y). Can someone with better math skills than myself explain why statement 2 is sufficient? Cheers!

-Ken


Mark my words bro, if you follow the way you are solving the questions you will face a difficult time with DS questions.
Always simply the equations till end, else you will complicate the things for you... - GOLDEN RULE.

now when you got y =x+y => x must be 0
now when you will put x=0 in both the statements you will have unique value of Y.
=> both are sufficient.

Now spend at least 15-20 mins on the advice I gave you and try to put in every question. Always simplify and then go ahead.
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Re: Tricky Algebra [#permalink] New post 16 Oct 2011, 11:44
even i mistakenly chose A...But its little difficult to understand why D?
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Re: Tricky Algebra [#permalink] New post 16 Oct 2011, 12:13
prateekbhatt wrote:
even i mistakenly chose A...But its little difficult to understand why D?


{x^2 - y^2} = (x+y)*(x-y)

The original expression ends up being: y = x + y thus x=0.

From here, both (1) and (2) can be solved:

(1) y=3
(2) -y=-3; y=3
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Re: Tricky Algebra [#permalink] New post 16 Oct 2011, 13:06
LGOdream wrote:
prateekbhatt wrote:
even i mistakenly chose A...But its little difficult to understand why D?


{x^2 - y^2} = (x+y)*(x-y)

The original expression ends up being: y = x + y thus x=0.

From here, both (1) and (2) can be solved:

(1) y=3
(2) -y=-3; y=3



Ohh got it.....I was thinking that if statement 1 is solving the purpose then why to check statement 2 ....i was wrong...

Thanks for solving.....
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Re: Tricky Algebra [#permalink] New post 22 Oct 2011, 11:13
+1 for D.

The given input is a good indication of what the questions expects (x not equal to y).

Simplifying the given equations we get:

x(x-y) = 0, since x -y = 0 or x is not equal to y, we know that x = 0.

Substituting x=0 in both conditions will give us the value of y. Hence D.
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink] New post 16 Dec 2011, 06:07
Excellent question kenman. Definitely worth another look. Its easy to miss out on the minor details and make careless mistakes as I did.
D is the right answer. Thanks for the explanation gurpreetsingh.
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the [#permalink] New post 03 Jan 2012, 21:45
D. From question stem we know

y = (x + y) (x-y) / (x-y)
y = (x + y) , y = ?

(1) x+ y = 3. so y = 3 SUFF.
(2) x -y = -3

x = -3 + y
Subs value of x in rephrased question stem

y = -3 + y + y
y = 3. SUFF.
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Re: If y = \frac {x^2 - y^2}{x-y} ( x \ne y ), then what is the   [#permalink] 03 Jan 2012, 21:45
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